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OpenStudy (amistre64):
well this ones tricker, you have to do more of a trial and error to break apart the "^3" cubic part.
OpenStudy (amistre64):
we can increase our odds by taking the factors of 10 as a pool of numbers to test.
1,10; 2,5
I use synthetic division to test it becasue its just less clutter
OpenStudy (amistre64):
lets try x = -1:
-1| 1 2 5 10
0 -1 -1 -4
-----------
1 1 4 6 <- not a zero, not a factor
OpenStudy (amistre64):
x = -2 maybe?
-2| 1 2 5 10
0 -2 0 -10
-----------
1 0 5 0 <- zero!! we have a winner..
OpenStudy (amistre64):
(x+2)(x^2+5x)
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